The ratio of three numbers is $$\frac{1}{4} : \frac{5}{6} : \frac{1}{3}$$ respectively. If the sum of the numbers is 68, what is the value of $$\frac{1}{3}$$ of the first number?

The ratio of three numbers is $$\frac{1}{4} : \frac{5}{6} : \frac{1}{3}$$ respectively.

Let's assume the three numbers are $$\frac{1}{4}y, \frac{5}{6}y, \frac{1}{3}y$$ respectively

If the sum of the numbers is 68.

$$\frac{1}{4}y+\frac{5}{6}y+\frac{1}{3}y = 68$$

$$\frac{3}{12}y+\frac{10}{12}y+\frac{4}{12}y=68$$

Value of $$\frac{1}{3}$$ of the first number = $$\frac{1}{3}$$ of $$\frac{1}{4}y$$

= $$\frac{1}{3}\times\frac{1}{4}\times\ 48$$

= $$\frac{1}{12}\times\ 48$$

= 4